Positive Solutions to a Generalized Second-Order Difference Equation with Summation Boundary Value Problem

Abstract

By using Krasnoselskii's fixed point theorem, we study the existence of positive solutions to the three-point summation boundary value problem Δ²𝑢(𝑡−1)+𝑎(𝑡)𝑓(𝑢(𝑡))=0, 𝑡∈{1,2,…,𝑇}, 𝑢(0)=𝛽∑^𝜂𝑠=1𝑢(𝑠), 𝑢(𝑇+1)=𝛼∑^𝜂𝑠=1𝑢(𝑠), where 𝑓 is continuous, 𝑇≥3 is a fixed positive integer, 𝜂∈{1,2,...,𝑇−1}, 0<𝛼<(2𝑇+2)/𝜂(𝜂+1), 0<𝛽<(2𝑇+2−𝛼𝜂(𝜂+1))/𝜂(2𝑇−𝜂+1), and Δ𝑢(𝑡−1)=𝑢(𝑡)−𝑢(𝑡−1). We show the existence of at least one positive solution if 𝑓 is either superlinear or sublinear.

1. Introduction

The study of the existence of solutions of multipoint boundary value problems for linear second-order ordinary differential and difference equations was initiated by Ilin and Moiseev [1]. Then Gupta [2] studied three-point boundary value problems for nonlinear second-order ordinary differential equations. Since then, nonlinear second-order three-point boundary value problems have also been studied by many authors; one may see the text books [3, 4] and the papers [5–10]. However, all these papers are concerned with problems with three-point boundary condition restrictions on the difference of the solutions and the solutions themselves, for example, 𝑢𝑢(0(𝑢)0−()𝑢0−()𝛼𝑢0Δ=𝛼)(0Δ𝑢=0)𝑢(,=00(,)00=),=𝑢0(𝑎,0𝑇𝑢,𝑢+(𝑠()𝑇1𝑢)=+Δ(−𝑇𝑢𝑢1+𝑎()(𝑢𝑇𝑇=1(++)𝑠0)=11,=))𝛽,=𝑢𝑏,(0𝑠,),(1.1) and so forth.

In [5], Leggett-Williams developed a fixed point theorem to prove the existence of three positive solutions for Hammerstein integral equations. Since then, this theorem has been reported to be a successful technique for dealing with the existence of three solutions for the two-point boundary value problems of differential and difference equations; see [6, 7]. In [8], X. Lin and W. Liu, using the properties of the associate Green’s function and Leggett-Williams fixed point theorem, studied the existence of positive solutions of the problem.

In [9], Zhang and Medina studied the existence of positive solutions for second-order boundary value problems of difference equations by applying Krasnoselskii’s fixed point theorem. In [10], Henderson and Thompson used lower and upper solution methods to study the existence of multiple solutions for second-order discrete boundary value problems.

We are interested in the existence of positive solutions of the following second-order difference equation with three-point summation boundary value problem (BVP): Δ²𝑢𝑢(𝑡(0−)1=)𝛽+𝑎_𝜂(𝑡)𝑓(𝑢(𝑡))=0,𝑡∈{1,2,…,𝑇},_𝑠=1𝑢(𝑠),𝑢(𝑇+1)=𝛼^𝜂_𝑠=1𝑢(𝑠),(1.2) where 𝑓 is continuous, 𝑇≥3 is a fixed positive integer, 𝜂∈{1,2,…,𝑇−1}.

The aim of this paper is to give some results for existence of positive solutions to (1.2), assuming that 0<𝛼<(2𝑇+2)/𝜂(𝜂+1), 0<𝛽<(2𝑇+2−𝛼𝜂(𝜂+1))/𝜂(2𝑇−𝜂+1), and 𝑓 is either superlinear or sublinear. Set 𝑓₀=lim_𝑢→0+𝑓(𝑢)𝑢,𝑓^∞=lim_𝑢→∞𝑓(𝑢)𝑢.(1.3) Then 𝑓₀=0 and 𝑓_∞=∞ correspond to the superlinear case, and 𝑓₀=∞ and 𝑓_∞=0 correspond to the sublinear case. Let ℕ be the nonnegative integer; we let ℕ_𝑖,𝑗={𝑘∈ℕ|𝑖≤𝑘≤𝑗} and ℕ_𝑝=ℕ_0,𝑝. By the positive solution of (1.2), we mean that a function 𝑢(𝑡)∶ℕ_𝑇+1→[0,∞) and satisfies the problem (1.2).

Recently, Sitthiwirattham [11] proved the existence of positive solutions for the boundary value problem with summation condition Δ²𝑢(𝑡−𝑢1()0+)=𝑎(0𝑡,)𝑓(𝑢𝑢((𝑡𝑇))+=10),=𝑡𝛼∈_𝜂{1,2,…,𝑇},_𝑠=1𝑢(𝑠),(1.4) where 𝑓 is continuous, 𝑇≥3 is a fixed positive integer, 𝜂∈{1,2,…,𝑇−1}, and 0<𝛼<2𝑇+2/𝜂(𝜂+1).

Throughout this paper, we suppose the following conditions hold:(𝐴1)𝑓∈𝐶([0,∞),[0,∞));(𝐴2)𝑎∈𝐶(ℕ_𝑇+1,[0,∞)) and there exists 𝑡₀∈ℕ_𝜂,𝑇+1 such that 𝑎(𝑡₀)>0.

The proof of the main theorem is based upon an application of the following Krasnoselskii’s fixed point theorem in a cone.

Theorem 1.1 (see [12]). Let 𝐸 be a Banach space, and let 𝐾⊂𝐸 be a cone. Assume Ω₁, Ω₂ are open subsets of 𝐸 with 0∈Ω₁, Ω₁⊂Ω₂, and let 𝐴∶𝐾∩Ω₂⧵Ω₁⟶𝐾(1.5) be a completely continuous operator such that(i)‖𝐴𝑢‖⩽‖𝑢‖, 𝑢∈𝐾∩𝜕Ω₁, and ‖𝐴𝑢‖⩾‖𝑢‖, 𝑢∈𝐾∩𝜕Ω₂, or(ii)‖𝐴𝑢‖⩾‖𝑢‖, 𝑢∈𝐾∩𝜕Ω₁, and ‖𝐴𝑢‖⩽‖𝑢‖, 𝑢∈𝐾∩𝜕Ω₂.Then 𝐴 has a fixed point in 𝐾∩(Ω₂⧵Ω₁).

2. Preliminaries

We now state and prove several lemmas before stating our main results.

Lemma 2.1. Let 𝛽≠(2𝑇+2−𝛼𝜂(𝜂+1))/𝜂(2𝑇−𝜂+1). Then, for 𝑦∈𝐶(ℕ_𝑇+1,[0,∞)), the problem 𝑢(0)Δ=²𝑢𝛽(𝑡_𝜂−1)+𝑦(𝑡)=0,𝑡∈𝑁_1,𝑇,(2.1)_𝑠=1𝑢(𝑠),𝑢(𝑇+1)=𝛼^𝜂_𝑠=1𝑢(𝑠),(2.2) has a unique solution 𝑢(𝑡)=𝛽𝜂(𝜂+1)+2𝑡(1−𝛽𝜂)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1−𝛽(𝑇+1)−(𝛽−𝛼)𝑡(𝑇−𝑠+1)𝑦(𝑠)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝜂−1_𝑠=1−(𝜂−𝑠)(𝜂−𝑠−1)𝑦(𝑠)_𝑡−1_𝑠=1(𝑡−𝑠)𝑦(𝑠),𝑡∈ℕ_𝑇+1.(2.3)

Proof. From (2.1), we get Δ𝑢(Δ𝑡−𝑢Δ(1𝑡𝑢))−−(1)ΔΔ−𝑢𝑢((Δ𝑡𝑡𝑢−−(201)))===⋮−−−𝑦𝑦𝑦(((𝑡𝑡1))−,.1),(2.4)
We sum the above equations to obtainΔ𝑢(𝑡)=Δ𝑢(0)−^𝑡_𝑠=1𝑦(𝑠),𝑡∈ℕ_𝑇,(2.5) we denote ∑^{𝑞𝑠=𝑝}𝑦(𝑠)=0, if 𝑝>𝑞. Similarly, summing the above equation from 𝑡=0 to 𝑡=ℎ, we get 𝑢(ℎ+1)=𝑢(0)+(ℎ+1)Δ𝑢(0)−^ℎ_𝑠=1(ℎ+1−𝑠)𝑦(𝑠),ℎ∈ℕ_𝑇,(2.6) changing the variable from ℎ+1 to 𝑡, we have 𝑢(𝑡)=𝑢(0)+𝑡Δ𝑢(0)−^𝑡−1_𝑠=1(𝑡−𝑠)𝑦(𝑠)∶𝐴+𝐵𝑡−^𝑡−1_𝑠=1(𝑡−𝑠)𝑦(𝑠),𝑡∈ℕ_𝑇+1.(2.7) We sum (2.7) from 𝑠=1,2,…,𝜂, ^𝜂_𝑠=1𝑢(𝑠)=𝜂𝐴+𝜂(𝜂+1)2𝐵−^𝜂−1_{𝑠=1𝜂−𝑠𝑙=1}=𝜂𝐴+𝜂(𝜂+1)(𝑙−𝑠)𝑦(𝑠)2𝐵−21^𝜂−1_𝑠=1(𝜂−𝑠)(𝜂−𝑠+1)𝑦(𝑠).(2.8) By (2.2) from 𝑢(0)=𝛽∑^𝜂𝑠=1𝑢(𝑠), we get (1−𝛽𝜂)𝐴−𝛽𝜂(𝜂+1)2𝐵=−𝛽2^𝜂−1_𝑠=1(𝜂−𝑠)(𝜂−𝑠+1)𝑦(𝑠),(2.9) and from 𝑢(𝑇+1)=𝛼∑^𝜂𝑠=1𝑢(𝑠), we obtain (1−𝛼𝜂)𝐴+𝑇+1−𝛼𝜂(𝜂+1)2𝐵=^𝑇_𝑠=1(𝑇−𝑠+1)𝑦(𝑠)−𝛼2^𝜂−1_𝑠=1(𝜂−𝑠)(𝜂−𝑠+1)𝑦(𝑠).(2.10) Therefore, 𝐴=𝛽𝜂(𝜂+1)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1−𝛽(𝑇+1)(𝑇−𝑠+1)𝑦(𝑠)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝜂−1_𝑠=1𝐵=2(1−𝛽𝜂)(𝜂−𝑠)(𝜂−𝑠+1)𝑦(𝑠),(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1+𝛽−𝛼(𝑇−𝑠+1)𝑦(𝑠)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝜂−1_𝑠=1(𝜂−𝑠)(𝜂−𝑠+1)𝑦(𝑠).(2.11) Hence, (2.1)-(2.2) has a unique solution 𝑢(𝑡)=𝛽𝜂(𝜂+1)+2𝑡(1−𝛽𝜂)(2𝑡+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1−𝛽(𝑇+1)−(𝛽−𝛼)𝑡(𝑇−𝑠+1)𝑦(𝑠)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝜂−1_𝑠=1−(𝜂−𝑠)(𝜂−𝑠+1)𝑦(𝑠)_𝑡−1_𝑠=1(𝑡−𝑠)𝑦(𝑠),𝑡∈ℕ_𝑇+1.(2.12)

Lemma 2.2. Let 0<𝛼<(2𝑇+2)/𝜂(𝜂+1), 0<𝛽<(2𝑇+2−𝛼𝜂(𝜂+1))/𝜂(2𝑇−𝜂+1). If 𝑦∈𝐶(ℕ_𝑇+1,[0,∞)) and 𝑦(𝑡)⩾0 for 𝑡∈ℕ_1,𝑇, then the unique solution 𝑢 of (2.1)-(2.2) satisfies 𝑢(𝑡)⩾0 for 𝑡∈ℕ_𝑇+1.

Proof. From the fact that Δ²𝑢(𝑡−1)=𝑢(𝑡+1)−2𝑢(𝑡)+𝑢(𝑡−1)=−𝑦(𝑡)≤0, we know that 𝑢(𝑡)≥(𝑢(𝑡+1)+𝑢(𝑡−1))/2, so 𝑢(𝑡+1)/(𝑡+1)<𝑢(𝑡)/𝑡.
Hence𝑢(𝑇+1)−𝑢(0)𝑇+1<𝑢(𝜂𝜂)+−1𝑢(0),𝜂∈ℕ^1,𝑇,(2.13) since 𝑢(𝑇)⩾0 and 𝑢(0)≥0 imply that 𝑢(𝑡)⩾0 for 𝑡∈ℕ_𝑇+1.
Moreover, from 𝑢(𝑖)>(𝑖/𝜂)𝑢(𝜂)+((𝜂−𝑖)/𝜂)𝑢(0), we get^𝜂_𝑠=1𝑢(𝑠)>1=1𝜂𝑢(𝜂)+𝜂−𝜂1𝑢(0)+2𝜂𝑢(𝜂)+𝜂−𝜂2𝑢(0)+⋯+𝜂𝜂𝑢(𝜂)+𝜂−𝜂𝜂𝑢(0)=𝜂1𝑢(𝜂)[1+2+⋯+𝜂]+1𝜂𝑢(0)[(𝜂−1)+(𝜂−2)+⋯+0]𝜂𝑢(𝜂)12𝜂(𝜂+1)+𝜂1𝑢(0)𝜂²−1=12𝜂(𝜂+1)2(𝜂+1)𝑢(𝜂)+21(𝜂−1)𝑢(0).(2.14) Combining (2.14) with (2.2), we can get 𝑢(0)>𝛽(𝜂+1)2−𝛽(𝜂−1)𝑢(𝜂),(2.15) again combining (2.2), (2.14), and (2.15), we obtain 𝑢(𝑇+1)>𝛼(𝜂+1)2−𝛽(𝜂−1)𝑢(𝜂),(2.16) such that 2−𝛽(𝜂−1)>2−𝛽𝜂>2−2𝑇+2−𝛼𝜂(𝜂+1)2𝑇−𝜂+1=2(𝑇−2𝜂𝑇)−+𝜂𝛼+𝜂(1𝜂+1)>0.(2.17) By using (2.13), (2.15), and (2.16), we obtain 2−2𝛽𝜂𝜂𝑢(𝜂)⩾(𝛼−𝑇𝛽)+(𝜂1+1)𝑢(𝜂).(2.18) If 𝑢(0)<0, then 𝑢(𝜂)<0. It implies that (2𝑇+2−𝛼𝜂(𝜂+1))/𝜂(2𝑇−𝜂+1)⩽𝛽, a contradiction to 𝛽<(2𝑇+2−𝛼𝜂(𝜂+1))/𝜂(2𝑇−𝜂+1). If 𝑢(𝑇)<0, then 𝑢(𝜂)<0, and the same contradiction emerges. Thus, it is true that 𝑢(0)⩾0, 𝑢(𝑇)⩾0, together with (2.13), we have 𝑢(𝑡)⩾0,𝑡∈ℕ_𝑇+1.(2.19) This proof is complete.

Lemma 2.3. Let 𝛼𝜂(𝜂+1)>2𝑇+2, 𝛽>max{(2𝑇+2−𝛼𝜂(𝜂+1))/𝜂(2𝑇−𝜂+1),0}. If 𝑦∈𝐶(ℕ_𝑇+1,[0,∞)) and 𝑦(𝑡)⩾0 for 𝑡∈ℕ_1,𝑇, then problem (2.1)-(2.2) has no positive solutions.

Proof. Suppose that problem (2.1)-(2.2) has a positive solution 𝑢 satisfying 𝑢(𝑡)⩾0, 𝑡∈ℕ_𝑇+1, and there is a 𝜏₀∈ℕ_1,𝑇 such that 𝑢(𝜏₀)>0.
If 𝑢(𝑇+1)>0, then ∑^𝜂𝑠=1𝑢(𝑠)>0. It implies𝑢(0)=𝛽^𝜂_𝑠=1𝑢(𝑠)>2𝑇+2−𝛼𝜂(𝜂+1)𝜂(2𝑇−𝜂+1)^𝜂_𝑠=1⩾𝜂(𝑇+1)(𝑢(0)+𝑢(𝜂))𝑢−(𝑠𝜂)(𝜂+1)𝑢(𝑇+1)𝜂(2𝑇−𝜂+1),(2.20) that is, 𝑢(𝑇+1)−𝑢(0)𝑇+1>𝑢(𝜂𝜂)+−1𝑢(0),(2.21) which is a contradiction to (2.13).
If 𝑢(𝑇+1)=0, then ∑^𝜂𝑠=1𝑢(𝑠)𝑑𝑠=0. When 𝜏₀∈ℕ_1,𝜂−1, we get 𝑢(𝜏₀)>𝑢(𝑇)=0>𝑢(𝜂), which contradicts to (2.13). When 𝜏₀∈ℕ_𝜂+1,𝑇, we get 𝑢(𝜂)⩽0=𝑢(0)<𝑢(𝜏₀), which contradicts to (2.13) again. Therefore, no positive solutions exist.

Let 𝐸=𝐶(ℕ_𝑇+1,[0,∞)), then 𝐸 is a Banach space with respect to the norm ‖𝑢‖=sup_{𝑡∈ℕ𝑇+1}||𝑢(𝑡)||.(2.22)

Lemma 2.4. Let 0<𝛼<(2𝑇+2)/𝜂(𝜂+1), 0<𝛽<(2𝑇+2−𝛼𝜂(𝜂+1))/𝜂(2𝑇−𝜂+1). If 𝑦∈𝐶(ℕ_𝑇+1,[0,∞)) and 𝑦(𝑡)⩾0, then the unique solution to problem (2.1)-(2.2) satisfies inf_{𝑡∈ℕ𝑇+1}𝑢(𝑡)⩾𝛾‖𝑢‖,(2.23) where 𝛾∶=min𝛼(𝜂+1)(𝑇+1−𝜂)𝛽((𝜂𝑇++11))((𝑇2+−1𝛽−(𝜂𝜂−)1))−𝛼𝜂(𝜂+1),(2−𝛽𝛼(𝜂𝜂(−𝜂1+))1()𝑇+1),(2−𝛽(𝜂−1))(𝑇+1),(2−𝛽𝛽(𝜂𝜂(−𝜂1+))1()𝑇+1).(2.24)

Proof. Let 𝑢(𝑡) be maximal at 𝑡=𝜏₁, when 𝜏₁∈ℕ_1,𝑇 and ‖𝑢‖=𝑢(𝜏₁). We divide the proof into two cases.
Case i. If 𝑢(0)⩾𝑢(𝑇+1) and inf_{𝑡∈ℕ𝑇+1}𝑢(𝑡)=𝑢(𝑇+1), then either 0⩽𝜏₁⩽𝜂<𝑇+1 or 0<𝜂<𝜏₁<𝑇+1, if 0⩽𝜏₁⩽𝜂<𝑇+1, then 𝑢𝜏₁⩽𝑢(𝑇+1)+𝑢(𝑇+1)−𝑢(𝜂)𝑇+1−𝜂𝜏¹⩽𝑢(𝑇+1)+𝑢(𝑇+1)−𝑢(𝜂)−(𝑇+1)⩽𝑢(𝑇+1)1−𝑇(+𝑇1+−1)𝜂−(𝑇(+0−1)((𝑇2+−1𝛽))(𝜂−1))/𝛼(𝜂+1)⩽𝑢(𝑇+1)(𝑇+1)(2−𝛽(𝜂−1))𝑇−+𝛼𝜂1(−𝜂+𝜂1)𝛼(𝑇+1)(𝑇+1−𝜂).(2.25) This implies inf_{𝑡∈ℕ𝑇+1}𝑢(𝑡)⩾𝛼(𝑇+1)(𝑇+1−𝜂)(𝑇+1)(2−𝛽(𝜂−1))−𝛼𝜂(𝜂+1)‖𝑢‖.(2.26)
Similarly, if 0<𝜂<𝜏₁<𝑇+1, from𝑢(𝜂)𝜂⩾𝑢𝜏¹𝜏₁⩾𝑢𝜏¹𝑇+1,(2.27) together with (2.16), we have 𝑢(𝑇+1)⩾𝛼𝜂(𝜂+1)(2−𝛽(𝜂−1))(𝑇+1)𝑢𝜏¹.(2.28) This implies inf_{𝑡∈ℕ𝑇+1}𝑢(𝑡)⩾𝛼𝜂(𝜂+1)(2−𝛽(𝜂−1))(𝑇+1)‖𝑢‖.(2.29)
Case ii. If 𝑢(0)⩽𝑢(𝑇+1) and inf_{𝑡∈ℕ𝑇+1}𝑢(𝑡)=𝑢(0), then either 0<𝜏₁<𝜂<𝑇+1 or 0<𝜂⩽𝜏₁⩽𝑇+1, by (2.13). If 0<𝜏₁<𝜂<𝑇+1, from 𝑢(𝜂)𝑇+1−𝜂⩾𝑢𝜏¹𝑇+1−𝜏₁⩾𝑢𝜏¹𝑇+1,(2.30) together with (2.15), we have 𝑢(0)>𝛽(𝜂+1)(𝑇+1−𝜂)(2−𝛽(𝜂−1))(𝑇+1)𝑢𝜏¹.(2.31) Hence, inf_{𝑡∈ℕ𝑇+1}𝑢(𝑡)>𝛽(𝜂+1)(𝑇+1−𝜂)(2−𝛽(𝜂−1))(𝑇+1)‖𝑢‖.(2.32) If 0<𝜂⩽𝜏₁⩽𝑇+1, from 𝑢𝜏₁𝑇+1⩽𝑢𝜏¹𝜏₁⩽𝑢(𝜂)𝜂,(2.33) together with (2.15), we have 𝑢(0)<𝛽𝜂(𝜂+1)(2−𝛽(𝜂−1))(𝑇+1)𝑢𝜏¹.(2.34) This implies inf_{𝑡∈𝑁𝑇+1}𝑢(𝑡)<𝛽𝜂(𝜂+1)(2−𝛽(𝜂−1))(𝑇+1)‖𝑢‖.(2.35) This completes the proof.

In the rest of the paper, we assume that 0<𝛼<(2𝑇+2)/𝜂(𝜂+1), 𝑇∈ℕ_1,𝑇; 0<𝛽<(2𝑇+2−𝛼𝜂(𝜂+1))/𝜂(2𝑇−𝜂+1). It is easy to see that the BVP (1.2) has a solution 𝑢=𝑢(𝑡) if and only if 𝑢 is a solution of the operator equation 𝐴𝑢(𝑡)≜𝛽𝜂(𝜂+1)+2𝑡(1−𝛽𝜂)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1−𝛽(𝑇+1)−(𝛽−𝛼)𝑡(𝑇−𝑠+1)𝑢(𝑠)𝑓(𝑢(𝑠))(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝜂−1_𝑠=1−(𝜂−𝑠)(𝜂−𝑠+1)𝑢(𝑠)𝑓(𝑢(𝑠))_𝑡−1_𝑠=1(𝑡−𝑠)𝑢(𝑠)𝑓(𝑢(𝑠)).(2.36)Denote 𝐾=𝑢∈𝐸∶𝑢⩾0,min_{𝑡∈ℕ𝑇+1}𝑢(𝑡)⩾𝛾‖𝑢‖,(2.37)

where 𝛾 is defined in (2.24).

It is obvious that 𝐾 is a cone in 𝐸. Since 𝐴𝑢=𝑢 and from Lemmas 2.2 and 2.4, then 𝐴(𝐾)⊂𝐾. It is also easy to check that 𝐴∶𝐾→𝐾 is completely continuous. In the following, for the sake of convenience, setΛ₁=(2𝑇+2)(1−𝛽𝜂)+𝛽𝜂(𝜂+1)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1Λ(𝑇−𝑠+1)𝑎(𝑠),₂=𝛾(2−𝛽𝜂+𝛽)(𝑇−𝜂)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1𝑠𝑎(𝑠).(2.38)

3. Main Results

Now we are in the position to establish the main result.

Theorem 3.1. The BVP (1.2) has at least one positive solution in the case(𝐻₁)𝑓₀=0 and 𝑓_∞=∞ (superlinear) or(𝐻₂)𝑓₀=∞ and 𝑓_∞=0 (sublinear).

Proof. Superlinear Case
Let (𝐻₁) hold. Since 𝑓₀=lim_𝑢→0⁺(𝑓(𝑢)/𝑢)=0 for any 𝜀∈(0,Λ¹⁻¹], there exists 𝜌_∗ such that 𝑓(𝑢)⩽𝜀𝑢for𝑢∈0,𝜌_∗.(3.1) Let Ω_𝜌∗={𝑢∈𝐸∶‖𝑢‖<𝜌_∗} for any 𝑢∈𝐾∩𝜕Ω_𝜌∗. From (3.1), we get 𝐴𝑢(𝑡)=2𝑡(1−𝛽𝜂)+𝛽𝜂(𝜂+1)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1−𝛽(𝑇+1)−(𝛽−𝛼)𝑡(𝑇−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝜂−1_𝑠=1−(𝜂−𝑠)(𝜂−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))_𝑡−1_𝑠=1⩽(𝑡−2𝑡𝑠()1𝑎−(𝑠𝛽)𝜂𝑓)(+𝑢(𝛽𝑠𝜂))(𝜂+1)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1⩽2(𝑇+1)(1−𝛽𝜂)+𝛽𝜂(𝜂+1)(𝑇−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1⩽𝜀𝜌_∗(𝑇−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))(2𝑇+2)(1−𝛽𝜂)+𝛽𝜂(𝜂+1)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1=𝜀Λ₁(𝑇−𝑠+1)𝑎(𝑠)𝜌_∗⩽𝜌_∗=‖𝑢‖,(3.2) which yields ‖𝐴𝑢‖⩽‖𝑢‖for𝑢∈𝐾∩𝜕Ω_𝜌∗.(3.3) Further, since 𝑓_∞=lim_𝑢→∞(𝑓(𝑢)/𝑢)=∞, then, for any 𝑀^∗∈[Λ²⁻¹,∞), there exists 𝜌^∗>𝜌_∗ such that 𝑓(𝑢)⩾𝑀^∗𝑢for𝑢⩾𝛾𝜌^∗.(3.4) Set Ω_𝜌∗={𝑢∈𝐸∶‖𝑢‖<𝜌^∗} for 𝑢∈𝐾∩𝜕Ω_𝜌∗.
Since 𝑢∈𝐾, min_{𝑡∈𝑁𝑇}𝑢(𝑡)⩾𝛾‖𝑢‖=𝛾𝜌^∗. Hence, for any 𝑢∈𝐾∩Ω_𝜌∗, from (3.4) and (2.23), we get 𝐴𝑢(𝜂)=(2−𝛽𝜂+𝛽)𝜂(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1−𝛽(𝑇+1)−(𝛽−𝛼)𝜂(𝑇−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝜂_𝑠=1−(𝜂−𝑠)(𝜂−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))_𝜂−1_𝑠=1=(𝜂−𝑠)𝑎(2(𝑠−)𝑓𝛽𝜂(𝑢+(𝑠𝛽)))𝜂(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1+1(𝑇−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))×(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)_𝜂−1_𝑠=1=(𝜂−𝑠)[(−2(−2𝛽−𝜂𝛽+𝜂𝛽+)𝛽𝜂)𝑇+(𝛽(𝑇−𝜂)+𝛼𝜂+1)𝑠+(𝜂−1)𝛽]𝑎(𝑠)𝑓(𝑢(𝑠))(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1−𝑇(2−𝛽𝜂+𝛽)(𝑇−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝜂−1_𝑠=1+𝛽(𝑡−𝜂)+𝛼𝜂+1(𝜂−𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))+(2𝑇+2−𝛼𝜂(𝜂(+𝜂−1)1))−𝛽𝛽𝜂(2𝑇−𝜂+1)^𝜂−1_𝑠=1𝜂𝑠−𝑠²𝑎(𝑠)𝑓(𝑢(𝑠))(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝜂−1_𝑠=1⩾(2−𝛽𝜂+𝛽)𝜂(𝜂−𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1−𝑇(2−𝛽𝜂+𝛽)(𝑇−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1=(2−𝛽𝜂+𝛽)(𝑇−𝜂)(𝜂−𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_s=1⩾𝛾𝜌_∗𝑠𝑎(𝑠)𝑓(𝑢(𝑠))𝑀^∗(2−𝛽𝜂+𝛽)(𝑇−𝜂)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1𝑠𝑎(𝑠)=𝑀^∗Λ₂𝜌^∗⩾𝜌^∗=‖𝑢‖,(3.5) which implies ‖𝐴𝑢‖⩾‖𝑢‖for𝑢∈𝐾∩𝜕Ω_𝜌∗.(3.6)
Therefore, from (3.3), (3.6), and Theorem 1.1, it follows that 𝐴 has a fixed point in 𝐾∩(Ω_𝜌∗⧵Ω_𝜌∗) such that 𝜌_∗⩽‖𝑢‖⩽𝜌^∗.
Sublinear Case
Let (𝐻₂) hold. In view of 𝑓₀=lim_𝑢→0⁺(𝑓(𝑢)/𝑢)=∞ for any 𝑀_∗∈[Λ²⁻¹,∞), there exists 𝑟_∗>0 such that 𝑓(𝑢)⩾𝑀_∗𝑢for0⩽𝑢⩽𝑟_∗.(3.7) Set Ω_𝑟∗={𝑢∈𝐸∶‖𝑢‖<𝑟_∗} for 𝑢∈𝐾∩𝜕Ω_𝑟∗. Since 𝑢∈𝐾, then min_{𝑡∈ℕ𝑇+1}𝑢(𝑡)⩾𝛾‖𝑢‖=𝛾𝑟_∗. Thus, from (3.7) for any 𝑢∈𝐾∩𝜕Ω_𝑟∗, we can get 𝐴𝑢(𝜂)=(2−𝛽𝜂+𝛽)𝜂(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1−𝛽(𝑇+1)−(𝛽−𝛼)𝜂(𝑇−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝜂−1_𝑠=1−(𝜂−𝑠)(𝜂−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))_𝜂−1_𝑠=1⩾𝛾𝑟_∗(𝜂−𝑠)𝑎(𝑠)𝑓(𝑢(𝑠))𝑀_∗(2−𝛽𝜂+𝛽)(𝑇−𝜂)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1𝑠𝑎(𝑠)=𝑀_∗Λ₂𝑟_∗⩾𝑟_∗=‖𝑢‖,(3.8) which yields ‖𝐴𝑢‖⩾‖𝑢‖for𝑢∈𝐾∩𝜕Ω_𝑟∗.(3.9)
Since 𝑓_∞=lim_𝑢→∞(𝑓(𝑢)/𝑢)=0, then, for any 𝜀₁∈(0,Λ¹⁻¹], there exists 𝑟₀>𝑟_∗ such that𝑓(𝑢)⩽𝜀₁𝑢for𝑢∈𝑟₀,∞.(3.10) We have the following two cases.
Case i. Suppose that 𝑓(𝑢) is unbounded, then, from 𝑓∈𝐶([0,∞),[0,∞)), we know that there is 𝑟^∗>𝑟₀ such that 𝑓(𝑢)⩽𝑓𝑟^∗for𝑢∈0,𝑟^∗.(3.11) Since 𝑟^∗>𝑟₀, then, from (3.10) and (3.11), one has 𝑓(𝑢)⩽𝑓𝑟^∗⩽𝜀₁𝑟^∗for𝑢∈0,𝑟^∗.(3.12) For 𝑢∈𝐾, ‖𝑢‖=𝑟^∗, from (3.12), we obtain 𝐴𝑢(𝑡)⩽(2𝑇+2)(1−𝛽𝜂)+𝛽𝜂(𝜂+1)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1⩽𝜀₁(𝑇−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))𝑟^∗(2𝑇+2)(1−𝛽𝜂)+𝛽𝜂(𝜂+1)(2𝑇+2−𝛼(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1=𝜀₁(𝑇−𝑠+1)𝑎(𝑠)Λ₁𝑟^∗⩽𝑟^∗=‖𝑢‖.(3.13)
Case ii. Suppose that 𝑓(𝑢) is bounded, say 𝑓(𝑢)⩽𝑁 for all 𝑢∈[0,∞). Taking 𝑟^∗⩾max{𝑁/𝜀₁,𝑟_∗}, for 𝑢∈𝐾, ‖𝑢‖=𝑟^∗, we have 𝐴𝑢(𝑡)⩽(2𝑇+2)(1−𝛽𝜂)+𝛽𝜂(𝜂+1)(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1⩽𝑁(2𝑇+2)(1−𝛽𝜂)+𝛽𝜂(𝜂+1)(𝑇−𝑠+1)𝑎(𝑠)𝑓(𝑢(𝑠))(2𝑇+2−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1⩽𝜀₁(𝑇−𝑠+1)𝑎(𝑠)𝑟^∗(2𝑇+2)(1−𝛽𝜂)+𝛽𝜂(𝜂+1)(2𝑇−𝛼𝜂(𝜂+1))−𝛽𝜂(2𝑇−𝜂+1)^𝑇_𝑠=1=𝜀₁(𝑇−𝑠+1)𝑎(𝑠)Λ₁𝑟^∗⩽𝑟^∗=‖𝑢‖.(3.14) Hence, in either case, we may always set Ω_𝑟∗={𝑢∈𝐸∶‖𝑢‖<𝑟^∗} such that ‖𝐴𝑢‖⩽‖𝑢‖for𝑢∈𝐾∩𝜕Ω_𝑟∗.(3.15)
Hence, from (3.9), (3.15), and Theorem 1.1, it follows that 𝐴 has a fixed point in 𝐾∩(Ω_𝜌∗⧵Ω_𝜌∗) such that 𝑟_∗⩽‖𝑢‖⩽𝑟^∗. The proof is complete.

4. Some Examples

In this section, in order to illustrate our result, we consider some examples.

Example 4.1. Consider the BVP 𝑢(0Δ)²=𝑢(1t−1)+𝑡²𝑢^𝑘=0,𝑡∈𝑁_1,4,3²_𝑠=1𝑢(𝑠),𝑢(5)=23²_𝑠=1𝑢(𝑠).(4.1) Set 𝛼=2/3, 𝛽=1/3, 𝜂=2, 𝑇=4, 𝑎(𝑡)=𝑡², and 𝑓(𝑢)=𝑢^𝑘.
We can show that0<𝛼=23<53=𝜂2(𝑇𝜂++21),0<𝛽=13<37=2𝑇𝜂+(22𝑇−−𝛼𝜂𝜂+(𝜂1+)1).(4.2)
Case I. 𝑘∈(1,∞). In this case, 𝑓₀=0,𝑓_∞=∞, and 𝐻₁ of Theorem 3.1 holds. Then BVP (4.1) has at least one positive solution.
Case II. 𝑘∈(0,1). In this case, 𝑓₀=∞,𝑓_∞=0, and 𝐻₂ of Theorem 3.1 holds. Then BVP (4.1) has at least one positive solution.

Example 4.2. Consider the BVP Δ²𝑢(𝑡−1)+𝑒^𝑡𝑡^𝑒𝜋sin𝑢+2cos𝑢𝑢²=0,𝑡∈𝑁^1,4𝑢(0)=1,4³_𝑠=1𝑢(𝑠),𝑢(5)=13³_𝑠=1𝑢(𝑠).(4.3) Set 𝛼=1/3, 𝛽=1/4, 𝜂=3, 𝑇=4, 𝑎(𝑡)=𝑒^𝑡𝑡^𝑒, 𝑓(𝑢)=(𝜋sin𝑢+2cos𝑢)/𝑢².
We can show that0<𝛼=13<56=𝜂2(𝑇𝜂++21),0<𝛽=14<13=2𝑇𝜂+(22𝑇−−𝛼𝜂𝜂+(𝜂1+)1).(4.4)
Through a simple calculation we can get 𝑓₀=∞,𝑓_∞=0. Thus, by 𝐻₂ of Theorem 3.1, we can get BVP (4.3) has at least one positive solution.

Acknowledgment

This research is supported by the Centre of Excellence in Mathematics, Commission on Higher Education, Thailand.